local extremes

in Calculus Answers by Level 2 User (1.4k points)

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1 Answer

The given answer options are all wrong as can be seen from the graph below:

To confirm the error, we differentiate f(x):

f'(x)=4x³-12x=0 at extrema so 4x(x²-3)=0, giving us the points x=0, x=√3 and x=-√3.

f(0)=0, f(√3)=9-18=-9=f(-√3), minima at (-1.73,-9) and (1.73,-9), maximum at (0,0).

by Top Rated User (1.0m points)

Please inform your tutor of these unacceptable errors.

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